Question 38

The function $f(x) = 3 \sin x + 3 \cos x$ defined on the interval $(0, 2\pi)$ reaches its maximum value at $x =$ ....

Recall the derivative concepts for trigonometric functions

y = a \cdot \sin x \rightarrow y' = a \cdot \cos x

y = a \cdot \cos x \rightarrow y' = -a \cdot \sin x

The condition for maximum value is when $f'(x) = 0$ .

f'(x) = 0

3 \cos x - 3 \sin x = 0

3 \cos x = 3 \sin x

\frac{\sin x}{\cos x} = 1

\tan x = 1

Then the values of $x$ that satisfy $\tan x = 1$ are $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$ .

Check first which function has maximum and minimum values.

For the value $x = \frac{\pi}{4}$

x = \frac{\pi}{4} \rightarrow f\left(\frac{\pi}{4}\right) = 3 \sin \frac{\pi}{4} + 3 \cos \frac{\pi}{4}

x = \frac{\pi}{4} \rightarrow f\left(\frac{\pi}{4}\right) = 3\left(\frac{1}{2}\sqrt{2}\right) + 3\left(\frac{1}{2}\sqrt{2}\right)

x = \frac{\pi}{4} \rightarrow f\left(\frac{\pi}{4}\right) = 3\sqrt{2} \text{ (max)}

For the value $x = \frac{5\pi}{4}$

x = \frac{5\pi}{4} \rightarrow f\left(\frac{5\pi}{4}\right) = 3 \sin \frac{5\pi}{4} + 3 \cos \frac{5\pi}{4}

x = \frac{5\pi}{4} \rightarrow f\left(\frac{5\pi}{4}\right) = 3\left(-\frac{1}{2}\sqrt{2}\right) + 3\left(-\frac{1}{2}\sqrt{2}\right)

x = \frac{5\pi}{4} \rightarrow f\left(\frac{5\pi}{4}\right) = -3\sqrt{2} \text{ (min)}

Therefore, the maximum value is reached when $x = \frac{\pi}{4}$ .

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